Is the differential of left multiplication still left multiplication?

Question

Let $A$ be a matrix in the real n-dimensional general linear group $G$. Let $l: G \to G$, $l(B)=AB$ be left multiplication by $A$. Consider the differential of $l$ at the identity matrix $I$ of $G$, $$l_{\{*,I\}}: T_I(G) \to T_{l(I)}(G)$$

Obviously, $l(I)=A$. Therefore, the differential is $$l_{\{*,I\}}: T_I(G) \to T_A(G)$$

Is the differential $l_{*,I}$ of $l$, left multiplication by $A$, still left multiplication by $A$?

I think it is, and here is what I've tried:

Let $X_I \in T_I(G)$. We must show $l_{*,I}(X_I)(f) = AX_I(f)$ for all $f \in C^{\infty}_I(G)$.

(Strictly, $f: U \to \mathbb R$ is a smooth function for some $U$ containing $I$ and open in $G$ while the members of $C^{\infty}_I(G)$ are equivalence classes. I think the $C^{\infty}_I(G)$ here is treated as a collection of the representatives of the equivalence classes).

First, there exists a smooth a curve $s:(-\varepsilon,\varepsilon) \to G$ that starts at $I$ and whose velocity vector is $X_I$ for some $\varepsilon > 0$. This means $s(0)=I$ and for each $f$, we have $$X_I(f) = s'(0)(f) := s_{*,0}[\frac{d}{dt} \mid_0 f] := \frac{d}{dt} \mid_0 f(s(t))$$

Then, for each $f$, $$l_{*,I}(X_I f) = \frac{d}{dt} \mid_0 f(l(s(t)))$$

Now, $s(t)$ is a matrix for every $t$, so $l(s(t))=As(t)$. Therefore, the left hand side is:

$$l_{*,I}(X_I f) = \frac{d}{dt} \mid_0 f(As(t))$$

For the right hand side,

$$AX_I(f) = As'(0)(f) = A s_{*,0}[\frac{d}{dt} \mid_0 f] = A \frac{d}{dt} \mid_0 f(s(t))$$

By linearity of $X_I$, $AX_I(f) = X_I(Af)$. Hence,

$$A \frac{d}{dt} \mid_0 f(s(t)) = AX_I(f) = X_I(Af) = s'(0)(Af) = s_{*,0}[\frac{d}{dt} \mid_0 Af] = \frac{d}{dt} \mid_0 Af(s(t))$$

Therefore, the right hand side is

$$AX_I(f) = A \frac{d}{dt} \mid_0 f(s(t)) = \frac{d}{dt} \mid_0 Af(s(t))$$

If $Af(s(t)) = f(As(t))$, then why?

I think $f(s(t)$ and $f(As(t))$ are real numbers while $Af(s(t))$ is a real invertible matrix. I might have said something meaningless above. If everything is meaningful, then (3) below might be helpful, but please explain how exactly (3) is helpful.

If $Af(s(t)) \ne f(As(t))$, then how do I proceed instead, or where have I gone wrong?

I think the following are supposed to be helpful, if they are right. Please explain how any of them are helpful or wrong and if I have implicitly used any of them already.

$G$ is open in $\mathbb R^{n \times n}$. Then $T_A(G) \cong \mathbb R^{n \times n}$.
By (1), the differential is $$l_{\{*,I\}}: T_I(G) \to \mathbb R^{n \times n}$$
Also by (1), the tangent vector $X_I$, which is supposedly a function $X_I: C^{\infty}_I(G) \to \mathbb R$, is now (the isomorphic pre-image or image of) a constant real n by n matrix.
$\mathbb R^{n \times n}$ is isomorphic to $\mathbb R^{n^2}$.
By (4), we have for the coordinate chart $(\mathbb R^{n^2}, x^1, ..., x^{n^2})$ about $s(0)=I$, where $x^1, ..., x^{n^2}$ are the standard coordinates on $\mathbb R^{n^2}$, that a basis for $T_I(G)$ is $\{\frac{\partial}{\partial x^1}, ..., \frac{\partial}{\partial x^{n^2}}\}$, and hence,

$$X_I = s'(0) = \sum_{i=1}^{n^2} \dot{s}^i(t) \frac{\partial}{\partial x^i} \mid_{s(0)} = \sum_{i=1}^{n^2} \dot{s}^i(t) \frac{\partial}{\partial x^i} \mid_{I},$$

where $s^i = x^i \circ s$ is the ith component of $s$

Thanks in advance!

I think I have an answer:

$$Af(s(t)) \ne f(As(t))$$

but $$\frac{d}{dt} \mid_0 f(As(t)) = \frac{d}{dt} \mid_0 Af(s(t))$$

For the left hand side, we have

$$\frac{d}{dt} \mid_0 f(As(t)) = (As)_{*,0}[\frac{d}{dt} \mid_0](f)$$

For the right hand side, we have

$$\frac{d}{dt} \mid_0 Af(s(t)) = A \frac{d}{dt} \mid_0 f(s(t)) = As'(0)(f) = A(s_{*,0}[\frac{d}{dt} \mid_0])(f)$$

It is correct and in this case it is simple identify the algebra ok general linear group. — Federico Fallucca, Jan 15 '19 at 07:59
@FedericoFallucca The 'algebra' of the general linear group is what justifies $Af(s(t)) = f(As(t))$ ? What is the 'algebra' of a group? Is it group algebra ? Algebra over a field ? — Ekhin Taylor R. Wilson, Jan 15 '19 at 08:38
For the algebra I refer to Lie algebra related to the Lie Group G.. the differential of your map permits to identify the bracket of Lie algebra of G, that in this case is the space of matrices with the standard commutator. I’m seeing now that there is a problem.. your idea is correct but the identity $Af(s(t))=f(As(t))$ does not make sense because A is a matrix so $Af(s(t))$ is a matrix while $f(s(t))$ is a number. You can only proof that $dl_I(B)=AB$ that is equivalent to say that $dl_I(X_I)(f)=AX_I(f)$ — Federico Fallucca, Jan 15 '19 at 09:23
@FedericoFallucca How do you prove the last statement without the concept of Lie algebras? — Ekhin Taylor R. Wilson, Jan 15 '19 at 10:01
The general linear group $GL(n)$ is an open sub-manifold of $M=M(n, n)=\mathbb{R}^{n^2}$. So they are locally diffeomorphic hence we can differentiate like we're in $M$. Note differential of a map is the induced linear map in the tangent space, while $M$ is a linear space, any tangent space is isomorphic to $M$, the differential of any linear map from $M$ to itself is the linear map itself (acting in the tangent space). — Xipan Xiao, Jan 15 '19 at 18:06
@XipanXiao How do you answer without local diffeomorphisms, sub-manifold or Lie algebras? — Ekhin Taylor R. Wilson, Jan 16 '19 at 04:40
@FedericoFallucca I think I proved the last statement in my most recent review. Is it right? — Ekhin Taylor R. Wilson, Jan 18 '19 at 07:41
@FedericoFallucca $(As){*,0}\frac{d}{dt} \mid_0 \ne A(s){*,0}\frac{d}{dt} \mid_0$ ? — Ekhin Taylor R. Wilson, Jan 28 '19 at 06:01

Matematleta · Answer 1 · 2023-02-06T02:28:21.230

I hope you'll pardon the necrobump. I think you are missing the gist of the exercise. As usual, there are identifications flying around. Lee, Proposition 3.13 has a very nice exposition of this. You don't need to know anything about Lie groups. If $X_B=\sum_{i,j=1}^nb_{ij}\frac{\partial}{\partial x^{ij}}$ with $B=(b_{ij})_{i=1}^n$ then you actually know what $s$ is. It is just $s(t)=I+Bt.$ And you want to prove that $l_*X_B=X_{AB}.$ So, calculate:

$(l_*X_B)f=l_*\circ s'(0)f=\frac{d}{dt}f(ls(0))=$

$\sum_{i,j=1}^n\dot{(l\circ s)}(0)\frac{\partial f}{\partial x^{ij}}(l(s(0)))=$

$\sum_{i,j=1}^n\left(\sum_{k=1}^na_{ik}b_{kj}\right)\frac{\partial f}{\partial x^{ij}}(l(s(0)))=$

$\sum_{i,j=1}^n\left(\sum_{k=1}^na_{ik}b_{kj}\right)\frac{\partial f}{\partial x^{ij}}(0\cdot AB)=$

$\sum_{i,j=1}^n\left(\sum_{k=1}^na_{ik}b_{kj}\right)\frac{\partial f}{\partial x^{ij}}(0)=X_{AB}f$

and so

$(l_*X_B)=X_{AB}.$

Is the differential of left multiplication still left multiplication?

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